Animated Solution for Mathematics - Vector Algebra: The unit vector perpendicular to the plane determined by $P(1,-1,2)$, $Q(2,0,-1)$ and $R(0,2,1)$ is .........

Defining Your Plane

Welcome, fellow traveler in the realm of three-dimensional space! Today, we are not just solving a problem; we are sculpting a plane out of the void.

Imagine you are floating in a vast, empty coordinate system. You have three points: $P(1,-1,2)$, $Q(2,0,-1)$, and $R(0,2,1)$.

These points are like stars in the night sky. In geometry, any three non-collinear points act as a tripod, defining a perfectly flat, infinite surface—a plane.

Our mission is to find a 'normal'—a vector that stands perfectly upright, perpendicular to this surface. It is the ultimate test of your spatial intuition.

Building the Foundation

Before we can find the perpendicular, we must first understand the surface itself. We cannot simply 'see' the plane; we must define it using vectors.

Think of vectors as arrows that live on this plane. If we can find two such arrows, we have captured the essence of the plane's orientation.

Let us anchor ourselves at point $P$. We create our first vector, $\overrightarrow{PQ}$, by traveling from $P$ to $Q$. The math is simple subtraction: we subtract the coordinates of $P$ from $Q$.

Now, we need a second vector to span the plane. Let us travel from $P$ to $R$. Again, we subtract the coordinates of $P$ from $R$:

We now have two vectors, $\overrightarrow{PQ}$ and $\overrightarrow{PR}$, lying flat on our plane. We have successfully translated the geometry of points into the algebra of vectors.

This is the first step of every great JEE problem: translating the abstract into the concrete.

The Magic of the Cross Product

Now, we reach the heart of the problem. We need a vector that is perpendicular to both $\overrightarrow{PQ}$ and $\overrightarrow{PR}$.

In the toolkit of a mathematician, there is one tool designed specifically for this: the cross product. The cross product $\vec{n}=\overrightarrow{PQ}\times \overrightarrow{PR}$ is not just a calculation; it is a geometric generator.

It creates a new vector that is orthogonal to the plane formed by the original two. We set this up using a determinant:

Let us expand this carefully. For the $\hat{i}$ component, we ignore the first row and column, leaving us with $(1\times -1)-(-3\times 3)=-1+9=8$. So, we have $8\hat{i}$.

For the $\hat{j}$ component, we must remember the alternating sign rule. We ignore the first row and second column, leaving us with $(1\times -1)-(-3\times -1)=-1-3=-4$.

Applying the negative sign, we get $-(-4)\hat{j}=4\hat{j}$. Finally, for the $\hat{k}$ component, we ignore the first row and third column, leaving us with $(1\times 3)-(1\times -1)=3+1=4$.

Thus, we have $4\hat{k}$. Our normal vector is $\vec{n}=8\hat{i}+4\hat{j}+4\hat{k}$.

This vector is perpendicular to our plane, but it is not yet a unit vector. It is just a direction.

The Final Normalization

To make this a 'unit' vector, we must strip away its magnitude, leaving only the direction. The magnitude $|\vec{n}|$ is calculated using the Pythagorean theorem in 3D:

We can simplify $\sqrt{96}$ as $\sqrt{16\times 6}=4\sqrt{6}$. Now, we divide our vector $\vec{n}$ by its magnitude $4\sqrt{6}$.

But wait! Remember that a perpendicular vector can point in two directions—up or down. This is why we include the $\pm$ sign.

Dividing each term by 4, we arrive at our elegant final answer:

Look at that result. It is clean, precise, and perfectly aligned with the geometry of the problem.

You have taken three arbitrary points and derived a fundamental property of the plane they inhabit. This is the power of vector algebra. Keep practicing, keep visualizing, and never lose that curiosity!